Laplace transforms: the table, the shift rules and initial-value problems

Worked solutions

Problem 1

1. $\int_0^\infty e^{-st}e^{3t}\,dt = \int_0^\infty e^{-(s - 3)t}\,dt$Combine the exponentials.
2. $= \lim_{b\to\infty}\dfrac{1 - e^{-(s - 3)b}}{s - 3}$Replace $\infty$ by $b$, integrate, then take the limit.
3. $\mathcal{L}\{e^{3t}\} = \dfrac{1}{s - 3}$ for $s > 3$The exponential tends to $0$ only when $s - 3 > 0$; for $s \le 3$ the integral diverges.
4. $\int_0^\infty te^{-st}\,dt = \Big[-\dfrac{te^{-st}}{s}\Big]_0^\infty + \dfrac1s\int_0^\infty e^{-st}\,dt$Integration by parts with $u = t$, $dv = e^{-st}dt$. The bracket is $0$ for $s > 0$.
5. $\mathcal{L}\{e^{3t}\} = \frac{1}{s - 3}$ for $s > 3$ and $\mathcal{L}\{t\} = \frac{1}{s^2}$ for $s > 0$At $s = 4$ these are $1$ and $\tfrac{1}{16}$. The transform of $t$ is the transform of $1$, which is $\frac1s$, divided by $s$ once more.

Problem 2

1. $\mathcal{L}\{4\} = \dfrac4s$Four times $\mathcal{L}\{1\}$.
2. $\mathcal{L}\{t^2\} = \dfrac{2!}{s^3} = \dfrac{2}{s^3}$The table with $n = 2$; do not forget the factorial.
3. $\mathcal{L}\{\sin 2t\} = \dfrac{2}{s^2 + 4}$The table with $b = 2$: $b$ on top, $b^2$ below.
4. $F(s) = \dfrac4s - 3\cdot\dfrac{2}{s^3} + 5\cdot\dfrac{2}{s^2 + 4}$Linearity: the constants come through unchanged.
5. $F(s) = \frac4s - \frac{6}{s^3} + \frac{10}{s^2 + 4}$, for $s > 0$At $s = 1$ it is $4 - 6 + 2 = 0$, so $\int_0^\infty e^{-t}f(t)\,dt = 0$.

Problem 3

1. $\mathcal{L}\{\cos 3t\} = \dfrac{s}{s^2 + 9}$The table with $b = 3$.
2. $e^{-2t}$ means $a = -2$, so $s$ becomes $s + 2$$F(s - a)$ with $a = -2$ is $F(s + 2)$.
3. $\mathcal{L}\{e^{-2t}\cos 3t\} = \dfrac{s + 2}{(s + 2)^2 + 9}$Replace every $s$, in the numerator too.
4. $\mathcal{L}\{t^3\} = \dfrac{6}{s^4}$, and $e^{4t}$ turns $s$ into $s - 4$$3! = 6$.
5. $\mathcal{L}\{e^{-2t}\cos 3t\} = \frac{s + 2}{s^2 + 4s + 13}$ for $s > -2$, and $\mathcal{L}\{t^3e^{4t}\} = \frac{6}{(s - 4)^4}$ for $s > 4$The shift moves the region of convergence too. At $s = 1$ the first is $\tfrac{3}{18} = \tfrac16$; shifting the other way gives $-0.1$ (Mistake 3).

Problem 4

1. $\dfrac{5s + 1}{(s - 1)(s + 2)} = \dfrac{A}{s - 1} + \dfrac{B}{s + 2}$Two distinct linear factors.
2. $A = \dfrac{5 + 1}{1 + 2} = 2$Cover up $s - 1$ and set $s = 1$.
3. $B = \dfrac{-10 + 1}{-2 - 1} = 3$Cover up $s + 2$ and set $s = -2$.
4. $2(s + 2) + 3(s - 1) = 5s + 1$Check by recombining.
5. $f(t) = 2e^t + 3e^{-2t}$Each $\frac{1}{s - a}$ is $e^{at}$. Check: $f(0) = 5$, and $\lim_{s\to\infty} sF(s) = 5$ as well.

Problem 5

1. $s^2 + 4s + 13 = (s + 2)^2 + 9$Complete the square; the roots $-2 \pm 3i$ are complex, so there are no real linear factors.
2. $2s + 3 = 2(s + 2) - 1$Rewrite the numerator in terms of $s + 2$ as well.
3. $F(s) = \dfrac{2(s + 2)}{(s + 2)^2 + 9} - \dfrac13\cdot\dfrac{3}{(s + 2)^2 + 9}$A shifted cosine and a shifted sine, with the $3$ the sine needs on top.
4. $\dfrac{s + 2}{(s + 2)^2 + 9} \to e^{-2t}\cos 3t$, $\dfrac{3}{(s + 2)^2 + 9} \to e^{-2t}\sin 3t$The first shift theorem with $a = -2$, read backward.
5. $f(t) = 2e^{-2t}\cos 3t - \tfrac13e^{-2t}\sin 3t$A damped oscillation: decay rate $2$ from the real part of the roots, frequency $3$ from the imaginary part. Check: $f(0) = 2$, and $f'(0) = -4 - 1 = -5$.

Problem 6

1. $sY - 2 + 3Y = \dfrac{1}{s + 1}$Transform both sides; $\mathcal{L}\{y'\} = sY - y(0)$.
2. $Y = \dfrac{2}{s + 3} + \dfrac{1}{(s + 1)(s + 3)}$Collect $(s + 3)Y$ and divide.
3. $\dfrac{1}{(s + 1)(s + 3)} = \dfrac{1/2}{s + 1} - \dfrac{1/2}{s + 3}$Cover-up: $\tfrac12$ at $s = -1$ and $-\tfrac12$ at $s = -3$.
4. $Y = \dfrac{3/2}{s + 3} + \dfrac{1/2}{s + 1}$Combine the two terms in $\frac{1}{s + 3}$.
5. $y(t) = \tfrac32e^{-3t} + \tfrac12e^{-t}$Check: $y(0) = 2$, and $y' + 3y = -\tfrac92e^{-3t} - \tfrac12e^{-t} + \tfrac92e^{-3t} + \tfrac32e^{-t} = e^{-t}$.

Problem 7

1. $s^2Y - s - 2 + 3(sY - 1) + 2Y = \dfrac4s$$\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$ and $\mathcal{L}\{y'\} = sY - y(0)$.
2. $(s^2 + 3s + 2)Y = s + 5 + \dfrac4s$Move the initial-condition terms $s + 2 + 3$ to the right.
3. $Y = \dfrac{s^2 + 5s + 4}{s(s + 1)(s + 2)}$Put the right side over $s$ and divide by $(s + 1)(s + 2)$.
4. $Y = \dfrac{(s + 1)(s + 4)}{s(s + 1)(s + 2)} = \dfrac{s + 4}{s(s + 2)}$The factor $s + 1$ cancels, so the $e^{-t}$ mode never appears for these initial values.
5. $\dfrac{s + 4}{s(s + 2)} = \dfrac2s - \dfrac{1}{s + 2}$Cover-up: $2$ at $s = 0$ and $\frac{2}{-2} = -1$ at $s = -2$.
6. $y(t) = 2 - e^{-2t}$Check: $y(0) = 1$, $y'(0) = 2$, and $y'' + 3y' + 2y = -4e^{-2t} + 6e^{-2t} + 4 - 2e^{-2t} = 4$. Leaving out the initial conditions gives a solution with $y(0) = 0$ (Mistake 1).

Problem 8

1. $g(t) = u(t - 2)\,t$The unit step switches $t$ on at $t = 2$.
2. $\mathcal{L}\{u(t - 2)\,t\} = e^{-2s}\mathcal{L}\{t + 2\}$The second shift theorem needs the function written in terms of $t - 2$; equivalently, shift its argument by $+2$.
3. $\mathcal{L}\{t + 2\} = \dfrac{1}{s^2} + \dfrac2s$$t + 2$ is the same ramp, measured from the switch-on time: it starts at the value $2$.
4. $G(s) = e^{-2s}\Big(\dfrac{1}{s^2} + \dfrac2s\Big)$Multiply by $e^{-2s}$.
5. $G(s) = \frac{(1 + 2s)e^{-2s}}{s^2}$Check from the definition: $\int_2^\infty te^{-st}\,dt$ at $s = 1$ is $3e^{-2} \approx 0.406$, and $G(1) = 3e^{-2}$. Writing $e^{-2s}/s^2$ drops the jump to $2$ (Mistake 4).

Problem 9

1. $(s^2 + 4)Y = \dfrac{e^{-\pi s}}{s}$$\mathcal{L}\{u(t - \pi)\} = e^{-\pi s}/s$, and both initial values are $0$.
2. $Y = e^{-\pi s}\cdot\dfrac{1}{s(s^2 + 4)}$Keep the exponential aside; it only shifts the answer.
3. $\dfrac{1}{s(s^2 + 4)} = \dfrac14\Big(\dfrac1s - \dfrac{s}{s^2 + 4}\Big)$Partial fractions: $\frac{1}{4}$ at $s = 0$, and the rest is $\frac{Bs + C}{s^2 + 4}$ with $B = -\frac14$, $C = 0$.
4. $\dfrac{1}{s(s^2 + 4)} \to \tfrac14(1 - \cos 2t)$The table, without the shift.
5. $y(t) = \tfrac14u(t - \pi)\big(1 - \cos 2(t - \pi)\big) = \tfrac14u(t - \pi)(1 - \cos 2t)$, so $y(3\pi/2) = \tfrac12$The spring stays at rest until $t = \pi$, then oscillates between $0$ and $\tfrac12$ about its new equilibrium $\tfrac14$. Here $\cos 2(t - \pi) = \cos 2t$ because $2\pi$ is a full period; in general the shift must stay inside.

Problem 10

1. $(s^2 + 2s + 5)Y = F(s)$, so $H(s) = \dfrac{Y}{F} = \dfrac{1}{s^2 + 2s + 5}$At rest, the initial-condition terms are all $0$.
2. $h(t) = \mathcal{L}^{-1}\Big\{\dfrac{1}{(s + 1)^2 + 4}\Big\} = \tfrac12e^{-t}\sin 2t$$\mathcal{L}\{\delta\} = 1$, so $Y = H$. Complete the square and put the needed $2$ on top: $\frac12\cdot\frac{2}{(s + 1)^2 + 4}$.
3. Step: $Y = \dfrac{H(s)}{s}$, which is the transform of $\int_0^t h(\tau)\,d\tau$The convolution theorem: the response is $h * f$, and with $f = 1$ the convolution is the running integral of $h$.
4. $\dfrac{1}{s(s^2 + 2s + 5)} = \dfrac{1/5}{s} - \dfrac15\cdot\dfrac{s + 2}{s^2 + 2s + 5}$$A = \frac15$ at $s = 0$; then $B = -\frac15$ from the $s^2$ terms and $C = -\frac25$ from the $s$ terms.
5. $\dfrac{s + 2}{(s + 1)^2 + 4} = \dfrac{s + 1}{(s + 1)^2 + 4} + \dfrac12\cdot\dfrac{2}{(s + 1)^2 + 4}$Shift the numerator as well as the denominator.
6. $h(t) = \tfrac12e^{-t}\sin 2t$, and the step response is $\tfrac15 - \tfrac15e^{-t}\cos 2t - \tfrac{1}{10}e^{-t}\sin 2t$, which settles to $\tfrac15$The final-value theorem agrees: $\lim_{s\to0} s\cdot\frac{H(s)}{s} = H(0) = \frac15$, allowed because the poles $-1 \pm 2i$ are in the left half-plane. The response starts at $0$ with slope $0$, as a system at rest must.